Question

In $\Delta ABC,if{a}^4+b^4+c^4=2c^2(a^2+b^2)thenangle\angle C=$

• A. $\frac{\pi }{8}$ or $\frac{3\pi }{8}$
• B. $\frac{\pi }{12}$ or $\frac{5\pi }{12}$
• C. $\frac{\pi }{4}$ or $\frac{3\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\frac{\pi }{4}$ or $\frac{3\pi }{4}$

$a^4+b^4+c^4=2c^2(a^2+b^2)$
$a^4+b^4+c^4+2a^2{b}^2=2c^2(a^2+b^2)+2a^2{b}^2$

${(a^2+b^2-c^2)}^2=2a^2{b}^2$

${\left(\frac{a^2+b^2-c^2}{ab}\right)}^2=2$

$\frac{a^2+b^2-c^2}{ab}=\sqrt{2}or-\sqrt{2}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\cos C=\frac{\sqrt{2}}{2}$

$C=\frac{\pi }{4}$

$\cos C=-\frac{\sqrt{2}}{2}$

$C=3\frac{\pi }{4}$.